Question

An electric iron has a rating of 750W, 200V. Calculate :-

A) the current required [ans 3.75ohm ]

B),the resistance of its heating element (ans 53.3ohm)

C) energy consumed by the iron in 2 hours( ans 1.5 kwh)
I need calculation of the sums

Answer 1

Explanation:

A. P= V^2/R (R = V/i)

750 = (200)^2×i/200= 40000×i/200

200i= 750

i = 750/200 = 3.75ohm ans....

B. V^2/R = P

750 = (200)^2/R

R = 40000/750 = 53.3333333333 ohm

answer...

C. Energy consumed = W = Pt

W = 750×(2×60×60)=5400000j.

answer or in khW

1kwh = 3600000 then...

5400000/3600000 = 1.5kwH answer...

hope it helps you..... BRAINLIEST

Answer 2

Heya !!

_________________________________

GIVEN THAT :

Rating of iron = 750 W, 200 V

TO FIND :

( A ) The current required ?

We know,

Power = Potential Difference ( V ) × Current ( I )

=> 750 = 200 × Current

=> Current ( I ) = 750/200

=> Current ( I ) = 3.75 A

____________________________

( B ) The resistance of its heating element ?

We know,

Resistance ( R ) = V/I

{ By Ohm's law }

=> R = 200/3.75

=> R = 2000/375

=> Resistance ( R ) = 53.33 Ohm.

_____________________________

( C ) Energy consumed by the iron in 2 hours = ?

We know,

Power ( P ) = Energy/Time

where,

Power = 750 W

=> P = 0.75 kW

Time ( t ) = 2 hours

=> 0.75 = E/2

=> E = 0.75 × 2

=> Energy ( E ) = 1.5 kWh.

____________________________

THANKS !!